Question

If one zero of the polynomial$(a^2+9)x^2+13x+6a$ is reciprocal of the other, find the value of a

Answer 1

$\Rightarrow$ The given quadratic polynomial is $(a^2+9)x^2+13x+6a$.

$\Rightarrow$ Let one zero of the quadratic polynomial be $\alpha$.

$\therefore$ The other zero of the quadratic polynomial is $\frac{1}{\alpha }.$

$\Rightarrow$ $\alpha \times \frac{1}{\alpha }=\frac{6a}{a^2+9}$

$\Rightarrow$ $1=\frac{6a}{a^2+9}$

$\Rightarrow$ $a^2+9=6a$

$\Rightarrow$ $a^2-6a+9=0$

$\Rightarrow$ $(a-3{)}^2=0$

$\Rightarrow$ $a-3=0$

$\Rightarrow$ $a=3$

$\therefore$ The value of $a$ is $3$.